Assume the following problem: \begin{align} u_{tt}-\Delta u &= -u^3\quad \text{in}\ \Omega,\ t>0\\ u&=0 \quad \text{for}\ \partial \Omega \\ u(\cdot,0)&=u_0 \\ u_t(\cdot, 0)&=0 \end{align} It should be proven, that there exists at most one classical solution.
So assuming that $\omega =u,v$ are classical solutions, I multiply both sides with $\omega_t=(u-v)_t$ and do a partial integration, obtaining $$0= \int_{\Omega} \omega_{tt}\omega_t - \nabla \omega \nabla \omega_t +(u^3-v^3)(u-v)_t dx$$ Now with the product rule I get $$\int_{\Omega} \omega_{tt}\omega_t - \nabla \omega \nabla \omega_t dx=\frac{1}{2}\frac{d}{dt}\int_{\Omega} \omega_t^2 +|\nabla \omega|^2 dx$$
The nonlinear term is making trouble now. I want to get a nonnegative term containing (u-v) whose derivate after $t$ gives me $(u^3-v^3)(u-v)_t$. I played around a bit, but couldn't come up with anything useful.
I have seen tricks in similar problems like: $(u^3-v^3)(u-v)=(u-v)^2\frac{1}{2}(u^2+v^2+(u+v)^2)$, however the time derivate gives me trouble utilizing that.
Another approach I have seen is using the mean value theorem that gives us a $\xi(x,t) \in [min(u, v), max(u, v)]$, so that $(u^3-v^3)=\xi^2(u-v)$. However, since $\xi$ is dependant on time, I can't neither find nonnegative term.
Would be realy grateful if someone could enlighten me.